Question

A continuous-wave (cw) argon-ion laser beam has an average power of \(10.0 \mathrm{~W}\) and a beam diameter of \(1.00 \mathrm{~mm} .\) Assume that the intensity of the beam is the same throughout the cross section of the beam (which is not true, as the actual distribution of intensity is a Gaussian function). a) Calculate the intensity of the laser beam. Compare this with the average intensity of sunlight at Earth's surface ( \(1400 . \mathrm{W} / \mathrm{m}^{2}\) ). b) Find the root-mean-square electric field in the laser beam. c) Find the average value of the Poynting vector over time. d) If the wavelength of the laser beam is \(514.5 \mathrm{nm}\) in vacuum, write an expression for the instantaneous Poynting vector, where the instantaneous Poynting vector is zero at \(t=0\) and \(x=0\). e) Calculate the root-mean-square value of the magnetic field in the laser beam.

Short answer

Question: Calculate and compare the intensity of an argon-ion laser beam with a power of 10.0 W and a diameter of 1.00 mm to the intensity of sunlight. Then, find the root-mean-square values of the electric and magnetic fields and describe the Poynting vector in the laser beam. Answer: The intensity of the laser beam is approximately \(1.27 \times 10^7 \mathrm{W/m^2}\), which is much higher than the average intensity of sunlight at Earth's surface (\(1400 \mathrm{W/m^2}\)). The root-mean-square electric field in the laser beam is approximately \(4800 \mathrm{V/m}\), and the root-mean-square magnetic field is approximately \(1.59 \times 10^{-8} \mathrm{T}\). The average value of the Poynting vector over time is \(1.27 \times 10^7 \mathrm{W/m^2}\), and the expression for the instantaneous Poynting vector is given by \(\vec{S}(t, x) = \frac{1}{\mu_0} E_0 B_0 \sin^2(2 \pi (ct - x)/\lambda) \hat{i}\).

Step-by-step solution

a) Calculate the intensity of the laser beam and compare to sunlight intensity

First, let's calculate the intensity of the laser beam. Intensity (\(I\)) is defined as the power per unit area: $$ I = \frac{P}{A} $$ We are given the average power (\(P\)) as \(10.0 \mathrm{W}\). We can find the area (\(A\)) of the beam using the given diameter (\(d = 1.00 \mathrm{mm}\)). The area of the beam is in the shape of a circle, so the formula to calculate the area is: $$ A = \pi r^2 $$ Where \(r\) is the radius of the circular beam. The radius is half of the diameter, so \(r = 0.50 \mathrm{mm} = 0.0005 \mathrm{m}\). Now we can calculate the area: $$ A = \pi (0.0005)^2 \approx 7.85 \times 10^{-7} \mathrm{m^2} $$ Now we can find the intensity of the laser beam: $$ I = \frac{10.0 \mathrm{W}}{7.85 \times 10^{-7} \mathrm{m^2}} \approx 1.27 \times 10^7 \mathrm{W/m^2} $$ Comparing this with the average intensity of sunlight at Earth's surface (\(1400 \mathrm{W/m^2}\)), the laser beam's intensity is much higher.

b) Find the root-mean-square electric field in the laser beam

We can use the formula that relates the intensity of the laser beam with the root-mean-square electric field (\(E_{rms}\)) and the speed of light (\(c\)) in a vacuum: $$ I = \frac{1}{2} \epsilon_0 c E_{rms}^2 $$ Where \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\) is the vacuum permittivity. Solving for \(E_{rms}\), we get: $$ E_{rms} = \sqrt{\frac{2I}{\epsilon_0 c}} $$ Plugging in the known values for \(I\), \(\epsilon_0\), and \(c = 3.00 \times 10^8 \mathrm{m/s}\), we find the root-mean-square electric field: $$ E_{rms} = \sqrt{\frac{2(1.27 \times 10^7 \mathrm{W/m^2})}{(8.85 \times 10^{-12} \mathrm{F/m})(3.00 \times 10^8 \mathrm{m/s})}} \approx 4800 \mathrm{V/m} $$

c) Find the average value of the Poynting vector over time

The average value of the Poynting vector over time is the same as the intensity of the laser beam. Therefore, the average Poynting vector is: $$ \langle S \rangle = 1.27 \times 10^7 \mathrm{W/m^2} $$

d) Write an expression for the instantaneous Poynting vector

The instantaneous Poynting vector can be expressed as a function of time and position (\(x\)): $$ \vec{S}(t, x) = \frac{1}{\mu_0} \vec{E} \times \vec{B} $$ Where \(\mu_0 = 4 \pi \times 10^{-7} \mathrm{Tm/A}\) is the vacuum permeability, and \(\vec{E}\) and \(\vec{B}\) are the electric and magnetic fields, respectively. Since we are looking for an expression where the instantaneous Poynting vector is zero at \(t = 0\) and \(x = 0\), we can assume that the electric and magnetic fields are sinusoidal functions of time and position: $$ \vec{E}(t, x) = E_0 \sin(2 \pi (ct - x)/\lambda) \hat{k} $$ $$ \vec{B}(t, x) = B_0 \sin(2 \pi (ct - x)/\lambda) \hat{j} $$ Where \(\lambda = 514.5 \mathrm{nm}\) is the wavelength of the laser beam, and \(E_0\) and \(B_0\) are the amplitudes of the electric and magnetic fields, respectively. Using these expressions, we can find the instantaneous Poynting vector: $$ \vec{S}(t, x) = \frac{1}{\mu_0} E_0 B_0 \sin^2(2 \pi (ct - x)/\lambda) \hat{i} $$

e) Calculate the root-mean-square value of the magnetic field in the laser beam

Using the intensity formula, we can find the root-mean-square value of the magnetic field, \(B_{rms}\), by substituting \(\vec{B}\) in place of \(\vec{E}\): $$ I = \frac{1}{2} \mu_0 c B_{rms}^2 $$ Solving for \(B_{rms}\), we obtain: $$ B_{rms} = \sqrt{\frac{2I}{\mu_0 c}} $$ Plugging in the known values for \(I\), \(\mu_0\), and \(c\), we find the root-mean-square magnetic field: $$ B_{rms} = \sqrt{\frac{2(1.27 \times 10^7 \mathrm{W/m^2})}{(4 \pi \times 10^{-7} \mathrm{Tm/A})(3.00 \times 10^8 \mathrm{m/s})}} \approx 1.59 \times 10^{-8} \mathrm{T} $$